Momentum help

A billiard ball of mass [tex] m_{A}=0.400kg [/tex] moving with a speed [tex] v_{A} =1.8m/s[/tex] strikes a second ball, initially at rest, of mass [tex] M_{B}=0.500kg[/tex]. As a result of the collision, the first ball is deflected off at an angle of [tex]30\deg[/tex] with a speed of [tex]v'_{A}=1.1 m/s[/tex].

a) taking the x-axis as the positive direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the x and y directions seperatley.

B)Solve the equations for the speed, [tex] v'_{B}[/tex], and the angle, [tex]\theta'_{2}[/tex] of ball b. Do not assume the collision is elastic.

I tried my hand at it, but bear with me, I didn't have my calc handy so I couldn't simplify anything, so I had to carry a bunch of phrases. As far as answers, I got [tex]\theta_{2} = \sin^{-1} \frac{.5(1.1)\sin (\frac{\pi}{6})}{-.4(1.8)} [/tex] For Vfinal for the initially moving ball, I got...
[tex]V_f=\frac{.4(1.8) - .5(1.1)\cos(\frac{\pi}{6})}{.4\cos(\theta_{2})}[/tex]
Like I said though, I didn't have a good way to test it out or anything, but that it was I gots, where [tex]\theta_{2}[/tex] is in radians, and in standard position. The whole [tex]2\pi-\theta_2[/tex] thing was unaccurate because it will output [tex]\theta_2[/tex] in standard position or as a negative, and doing subtracting it from [tex]2\pi[/tex] will merely warp the results. Try it out and tell me if I was close or not.
[/tex] because the signs are different between the ref angle, which is what [tex]\theta_2[/tex] is, and the actual angle, which is the aforementioned angle.
To get, I basically plugged into [tex]m_1v_o=m_1v_{1f}\cos(\theta_{2})+m_2v_{2f}\cos(\fract{\pi}{6})[/tex]
and
[tex]0=m_1v_{1f}\sin(\theta_2)+m_2v_{2f}\sin(\frac{\pi}{6})[/tex]

Staff: Mentor

Originally posted by bard well im not sure whether this problem has a definite answer(as in 1.2 or 3.4 etc)or whther im suposed to solve in terms of variables

If they meant you to solve it in terms of variables, why did they bother giving you all those values? Just plug in the numbers, then rewrite the two equations. You'll have two (simple) equations with two unknowns.

I had made my classical error of using the reference angle for [tex]\theta[/tex] rather than the true angle, and I swapped a sign or two when I did it on paper. Sorry, at least when I reworked it my answers matched up :)